X-Plosives UVALive - 3644（并查集）

A secret service developed a new kind of explosive that attain its volatile property only when a specific

association of products occurs. Each product is a mix of two different simple compounds, to which we

call a binding pair. If N > 2, then mixing N different binding pairs containing N simple compounds

creates a powerful explosive. For example, the binding pairs A+B, B+C, A+C (three pairs, three

compounds) result in an explosive, while A+B, B+C, A+D (three pairs, four compounds) does not.

You are not a secret agent but only a guy in a delivery agency with one dangerous problem: receive

binding pairs in sequential order and place them in a cargo ship. However, you must avoid placing in

the same room an explosive association. So, after placing a set of pairs, if you receive one pair that

might produce an explosion with some of the pairs already in stock, you must refuse it, otherwise, you

must accept it.

An example. Lets assume you receive the following sequence: A+B, G+B, D+F, A+E, E+G,

F+H. You would accept the first four pairs but then refuse E+G since it would be possible to make the

following explosive with the previous pairs: A+B, G+B, A+E, E+G (4 pairs with 4 simple compounds).

Finally, you would accept the last pair, F+H.

Compute the number of refusals given a sequence of binding pairs.

Input

The input will contain several test cases, each of them as described below. Consecutive

test cases are separated by a single blank line.

Instead of letters we will use integers to represent compounds. The input contains several lines.

Each line (except the last) consists of two integers (each integer lies between 0 and 105

) separated by

a single space, representing a binding pair.

Each test case ends in a line with the number ‘-1’. You may assume that no repeated binding pairs

appears in the input.

Output

For each test case, the output must follow the description below.

A single line with the number of refusals.

Sample Input

1 2

3 4

3 5

3 1

2 3

4 1

2 6

6 5

-1

Sample Output

3

題目大意：有一些簡單點的化合物，每個化合物都有兩種元素組成，你是一個裝箱勞工，從實驗員那裡按照順序依次把一些簡單的化合物裝到車上。但是這裡存在一個安全隐患：如果車上存在k個化合物，正好包含k中元素，那麼他們将組成一個易爆混合物。為了安全起見，當你拿到一個化合物時，如果它和車上化合物形成易爆化合物，你應當拒絕撞車，程式設計輸出由多少個沒有撞車的化合物。

思路：

好像是一個裸題，先将元素并起來，如果并起來的元素他們根節點相同，說明車上已經有這兩種化合物了，再裝上車可能會形成易爆物，這時候計數一下，這個輸入格式有點無語。。。。

代碼：

#include<iostream>
#define MAXN 100005
using namespace std;

int a,b;
int parent[MAXN];
int ans=0;
void defin()
{
	for(int i=1;i<=MAXN;i++){
		parent[i]=i;
	}
}
int FIND_X(int x)
{
	if(x!=parent[x]){
		parent[x]=FIND_X(parent[x]);
	}
	return parent[x];
}
void UNION_S(int a,int b)
{
	int x=FIND_X(a);
	int y=FIND_X(b);
	    if(x!=y){
			parent[x]=y;
		}else{
			ans++;
		}
}
int main()
{
	while(cin>>a&&a!=-1)
	{
		defin();
		ans=0;
		cin>>b;
		UNION_S(a,b);
		while(cin>>a&&a!=-1){
			cin>>b;
			UNION_S(a,b);
		}
		cout<<ans<<endl;
	}
}